Let $X_1,…,X_n$ be independent, but not necessarily identical, non-negative random variables. Let $Z=\max_i(X_i)$. Fix a real $\tau > 0$. Is there a way to lower bound $$\mathbb{E}(Z|Z>\tau) > c \mathbb{E}(Z)$$ for constant $c>1$ (that can obviously depend on $\tau$). For example if $\tau$ is the median of $Z$ or $\mathbb{E}(Z)/2$
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$\begingroup$ Not in the $\tau = \mathbb{E}(Z)/2$ case since you might for example have the $X_i$ supported on $[6,10]$ so $\mathbb{E}(Z)/2 \le 5$ and $\mathbb{E}(Z\mid Z>\tau) =1 \times \mathbb{E}(Z)$. I am not sure what the $X_i$ contribute here rather than just looking at $Z$ $\endgroup$– HenryCommented Dec 12, 2022 at 8:50
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$\begingroup$ @Henry $X_i$ just here to restrict valid $Z$ variables (as max of non negative random variables) and also in case there is a lower bound in terms of mean/variance of the individual $X_i$ instead. $\endgroup$– AspiringMatCommented Dec 12, 2022 at 8:59
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$\begingroup$ In the median case I think you can say $\mathbb{E}(Z\mid Z>\tau)\ge \max(\tau,2\mathbb{E}(Z) -\tau)$ $\endgroup$– HenryCommented Dec 12, 2022 at 9:12
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$\begingroup$ @Henry can you sketch the proof for the median inequality? (Or in general the p-th percentile if you can generalize) $\endgroup$– AspiringMatCommented Dec 13, 2022 at 6:59
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1 Answer
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Requested in comments:
Clearly in general $\mathbb{E}[Z\mid Z\le \tau] \le \tau \lt \mathbb{E}[Z\mid Z>\tau]$, assuming the expectations exist.
If $\tau$ is the median of $Z$ then:
- $\mathbb P(Z>\tau) \le \frac12 \le \mathbb P(Z\le \tau)$
- $\mathbb{E}[Z] = \mathbb P(Z>\tau)\,\mathbb{E}[Z\mid Z>\tau] + \mathbb P(Z\le\tau)\,\mathbb{E}[Z\mid Z\le\tau]$
- so $\mathbb{E}[Z] \le \frac12 \mathbb{E}[Z\mid Z>\tau] + \frac12 \mathbb{E}[Z\mid Z\le\tau] $
- and thus $\mathbb{E}[Z] \le \frac12 \mathbb{E}[Z\mid Z>\tau]+ \frac12 \tau $
- implying $\mathbb{E}[Z\mid Z>\tau] \ge 2 \mathbb{E}[Z] -\tau$
- which combined with the earlier result gives $$\mathbb{E}(Z\mid Z>\tau)\ge \max(\tau,2\mathbb{E}(Z) -\tau).$$
If you extended this to the $p$th quantile of $Z$ so $\mathbb P(Z\le \tau) \ge p$ and $\mathbb P(Z>\tau)\le 1-p$ then the same argument would give you $\mathbb{E}[Z\mid Z>\tau] \ge \max\left(\tau,\frac{\mathbb{E}[Z]-p\tau}{1-p}\right). $
